Multiplying Binomials Calculator

Background.

Every product of two binomials comes down to four multiplications. In (ax + b)(cx + d), each term of the first bracket has to meet each term of the second: ax with cx, ax with d, b with cx, and b with d. Students learn the order as FOIL — First, Outer, Inner, Last — and teachers often draw it as a rectangle cut into four pieces whose areas add up to the whole. Either way the result is a quadratic: ac x² from the first pair, bd from the last, and a middle term built from the two cross products.

That middle term is where most mistakes happen, because it is the only place two pieces combine and the only place signs can cancel. In (x + 3)(x − 3) the outer and inner products are −3x and +3x, so the middle term vanishes and the answer is x² − 9. In (x + 3)² they are both +3x, and forgetting them produces the wrong x² + 9 instead of x² + 6x + 9.

Enter the four numbers that define the brackets — the coefficient of x and the constant for each — and the page returns the three coefficients of the expanded quadratic. Negative values and decimals are both allowed. It expands only: it does not factor, multiply longer polynomials or work with letters in place of the coefficients.

What is multiplying binomials calculator?

Multiplying two binomials means applying the distributive law twice so that the product (ax + b)(cx + d) is written as a single polynomial in standard form, ac x² + (ad + bc)x + bd. Here a and c are the coefficients of x in each bracket and b and d are the constant terms, all real numbers. The three outputs are the coefficient of x², the coefficient of x and the constant term of that expanded quadratic.

How to use this calculator.

1. 01Write each bracket as (number × x + constant). For (x − 4) that means 1 and −4; for (3 − 2x), reorder it as (−2x + 3), giving −2 and 3.
2. 02Enter the first bracket's values as a and b and the second bracket's as c and d. Fractions go in as decimals, such as 0.5 for ½.
3. 03Read the three coefficients and assemble the answer as (x² coefficient)x² + (x coefficient)x + constant, writing any negative coefficient as a subtraction.
4. 04To check by hand, substitute x = 1: the original brackets and the expanded quadratic must give the same value.

The formula.

Distributing the first bracket over the second gives ax(cx + d) + b(cx + d) = acx² + adx + bcx + bd. The two middle products, adx (outer) and bcx (inner), are like terms, so they add to (ad + bc)x. The result has three coefficients: ac for x², ad + bc for x and bd for the constant. Only the middle coefficient mixes the two brackets, which is why it alone can come out as zero, as in (x + 3)(x − 3) = x² − 9. A useful identity hides in the arithmetic: the outer and inner products multiply to the same value as the first and last, (ad)(bc) = (ac)(bd), and that is what the ac method of factoring exploits in reverse. If a or c is zero, one bracket is only a constant and the product is linear rather than quadratic.

A worked example.

Frequently asked questions.

References& sources.

1. [1]OpenStax, Intermediate Algebra 2e, Multiply Polynomials.
2. [2]OpenStax, College Algebra 2e, Polynomials.
3. [3]BIPM, The International System of Units (SI Brochure), 9th ed., version 3.01, coherent derived units and quantity equations.